University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

457

Pages:

5
Author:

Jabin F.

uniform magnetic field B m = Iabz F = ∫(I dl × B) = ∫(I dB(+y)) by right hand grip rule. AB F_AB = IAB(+y) (i) F_CD = IAB(-y) (ii) F_CD + F_AB = 0 Similarly F_BC = ∫(I dl × B) JC = ∫(I dB) = IB(+y) -b/2 F_DA = ∫(I dl × B) -b/2 = ∫(I dB(-y)) = IB(-y) F_DA + F_BC = 0 F_net = 0 F_net = 0 → Force by zero in uniform mag. field. # Torque F_net = 0 Torque = τ = r × F |F_OA| = |F_BC| = IbB τ = IbBsinθn̂ τ = mBsinθn̂ = m × B m = Iabn̂, B = B₀ẑ Ques: The magnetic fields in tesla in the regions separated by the z=0 plane are given by B₁=3x̂+5ẑ and B₂=x̂+3ŷ+5ẑ. The magnitude of the surface current density at the interface b/w the two regions is α×10⁶A/m Given the permeability of the free space μ₀=4π×10⁻⁷μ/A², The Value of α ....? Sol: B₁=3x̂+5ẑ-(A) B₂=x̂+3ŷ+5ẑ-(B) B₁-B₂=0 B₂"-B₁"=μ₀(k×n) [B₂x-B₁x]î + [B₂y-B₁y]ĵ = μ₀[(kxn̂i + kyĵ)×n̂] [B₂x-B₁x]î + [B₂y-B₁y]ĵ = -μ₀kxĵ + μ₀kyî Compare ⇒ B₂x-B₁x = μ₀ky ⇒ B₂y-B₁y = -μ₀kx Value from (A) and (B) ky = (1-3)/μ₀ ⇒ ky = -2/μ₀ kx = -(3-0)/(4πμ₀) ⇒ kx = -3/(4πμ₀) Ques: Far from the Earth, the Earth's magnetic field can be approximated as due to a bar magnet of magnetic pole strength 4x10¹⁴ Am. Assume this magnetic field is generated by a current carrying loop encircling the magnetic equator. The current required to do so is about 4x10^n A, where n is an integer. The value of n is ? [Earth circumference: 4x10⁷m] Soln # Pole strength m = IA H = I/2R => (I/2R) × πR² = Pole strength => (I/2) × πR = 4x10¹⁴ => (I/2) × 2x10⁷ = 4x10¹⁴ => I = 4x10⁷A = 4x10^n A